The average weight of all the 11 players of t | vedclass

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Question

(A) Total weight of all $11$ players $= 11 	imes 50 = 550 \, kg$.Sum of the weights of the first six lightest players $= 6 	imes 49 = 294 \, kg$.Sum

Vedclass · Accepted Answer

$56$

Vedclass · Answer

(A) Total weight of all $11$ players $= 11 	imes 50 = 550 \, kg$.Sum of the weights of the first six lightest players $= 6 	imes 49 = 294 \, kg$.Sum of the weights of the last six heaviest players $= 6 	imes 52 = 312 \, kg$.When we add the sum of the six lightest and the six heaviest players,the player at the $6^{th}$ position is counted twice.Weight of the $6^{th}$ player $= (Sum \, of \, first \, six) + (Sum \, of \, last \, six) - (Total \, weight \, of \, 11 \, players)$.Weight of the $6^{th}$ player $= 294 + 312 - 550 = 606 - 550 = 56 \, kg$.

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